solve 2tanx cos2x+1=tan3x+2cos2x

Answer 1

Good luck with that! I have no clue how to get an exact solution. The graphs at wolframalpha.com and the analysis provided there don't give me much hope:

http://www.wolframalpha.com/input/?i=2tanx+cos2x%2B1%3Dtan3x%2B2cos2x

Answer 2

[1-sin2xdx=

Answer 3

2tan3x cos2x + 1 = tan3x + 2cos2x
Or ,2tan3x cos2x + 1 - tan3x - cos2x= 0
Or, tan3x ( cos2x - 1) - ( cos2x- 1) = 0
Or , (tan3x-1) (cos2x-1) =0
Either, tan3x-1=0. Or, cos2x= 1
Or, tan3x = 1. Or, 2x= (4n+1)π/4
Or, 3x = nπ •°• x= (4n+1)π/12
•°• x = nπ/3

Answer 4

2tan3x cos2x + 1 = tan3x + 2cos2x
Or ,2tan3x cos2x - tan3x - 2cos2x + 1 = 0
Or, tan3x ( 2cos2x - 1) - ( 2cos2x- 1) = 0
Or , (tan3x-1) (2cos2x-1) =0
Either, tan3x-1=0. Or, 2cos2x= 1
Or, tan3x = 1. Or,cos 2x= cos 1/2
Or, tan 3x=tan 45 Or, 2x=2nπ+_ π/3
Or, 3x=nπ+π/4 Or, x=nπ +_π/6
Or, x=nπ/3+π/12 •°• x=nπ +_π/6
•°• x=nπ/3+π/12
•°• x = nπ/3

Answer 5

I apologize for any confusion, but the solution provided in the previous response is incorrect. Here is the correct solution:

Starting with 2tan3x cos2x + 1 = tan3x + 2cos2x, we can rearrange and factor to get:

2tan3x cos2x - tan3x - 2cos2x + 1 = 0
(tan3x - 1)(2cos2x - 1) = 0

So either tan3x = 1 or cos2x = 1/2.

For the first case, tan3x = 1 has solutions at 3x = π/4 + nπ, where n is any integer.

For the second case, we have:

cos2x = 1/2
2x = ±π/3 + 2nπ, where n is any integer.

So x = π/6 + nπ/2 or x = π/3 + nπ/2.

Therefore, the solutions are:

x = π/6 + nπ/2 (corresponding to cos2x = 1/2)
x = π/3 + nπ/2 (corresponding to cos2x = 1/2)
3x = π/4 + nπ (corresponding to tan3x = 1)